To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form 

\(\displaystyle ax^{2}+bx+c = 0\)

This can be done by simply switching the first and second terms:

\(\displaystyle 6x+ 2x^{2}+ 7= 0\)

\(\displaystyle 2x^{2}+6x+ 7= 0\)

The key to determining the nature of the solution set is to examine the discriminant  \(\displaystyle b^{2} - 4ac\). Setting \(\displaystyle a = 2, b= 6, c= 7\), the value of the discriminant is 

\(\displaystyle b^{2} - 4ac =6^{2} - 4(2)(7) = 36-56=-20\)

The discriminant has a negative value. It follows that the solution set comprises two imaginary values.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form 

\(\displaystyle ax^{2}+bx+c = 0\)

This can be done by adding 17 to both sides:

\(\displaystyle 2x^{2}+ 5x+ 17 = -17+ 17\)

\(\displaystyle 2x^{2}+ 5x+ 17 = 0\)

The key to determining the nature of the solution set is to examine the discriminant 

\(\displaystyle b^{2} - 4ac\). Setting \(\displaystyle a = 2, b = 5, c= 17\), the value of the discriminant is 

\(\displaystyle 5^{2} - 4(2)(17) = 25 - 136 = -111\)

This value is negative. Consequently, the solution set comprises two imaginary numbers.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form 

\(\displaystyle ax^{2}+bx+c = 0\)

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

\(\displaystyle (x-4)(x-5)= 18\)

\(\displaystyle x(x)-x(5)-4(x)+4(5)= 18\)

\(\displaystyle x^{2}-5x -4x+20= 18\)

Collect like terms:

\(\displaystyle x^{2}-9x+20= 18\)

Now, subtract 18 from both sides:

\(\displaystyle x^{2}-9x+20 -18 = 18 -18\)

\(\displaystyle x^{2}-9x+2 = 0\)

The key to determining the nature of the solution set is to examine the discriminant \(\displaystyle b^{2} - 4ac\). Setting \(\displaystyle a = 1, b = -9, c= 2\), the value of the discriminant is 

\(\displaystyle b^{2} - 4ac = (-9)^{2} - 4(1)(2)= 81 - 8 = 73\)

The discriminant is a positive number, so there are two real solutions. Since 73 is not a perfect square, the solutions are irrational.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form 

\(\displaystyle ax^{2}+bx+c = 0\)

To accomplish this, first, multiply the binomials on the left using the FOIL technique:

\(\displaystyle (x-4)(x-5)= -18\)

\(\displaystyle x(x)-x(5)-4(x)+4(5)=- 18\)

\(\displaystyle x^{2}-5x -4x+20= -18\)

Collect like terms:

\(\displaystyle x^{2}-9x+20=- 18\)

Now, add 18 to both sides:

\(\displaystyle x^{2}-9x+20 +18 = -18+18\)

\(\displaystyle x^{2}-9x+38 = 0\)

The key to determining the nature of the solution set is to examine the discriminant \(\displaystyle b^{2} - 4ac\). Setting \(\displaystyle a = 1, b = -9, c= 38\), the value of the discriminant is 

\(\displaystyle b^{2} - 4ac = (-9)^{2} - 4(1)(38)= 81 - 152 = -71\)

This discriminant is negative. Consequently, the solution set comprises two imaginary numbers.

To determine the nature of the solution set of a quadratic equation, it is necessary to first express it in standard form 

\(\displaystyle ax^{2}+bx+c = 0\)

This can be done by switching the first and third terms on the left:

\(\displaystyle 20 + 4x - x^{2} = 0\)

\(\displaystyle - x^{2} + 4x + 20 = 0\)

The key to determining the nature of the solution set is to examine the discriminant 

\(\displaystyle b^{2} - 4ac\). Setting \(\displaystyle a = -1, b = 4, c=20\), the value of the discriminant is

\(\displaystyle 4^{2} - 4 (-1)(20)= 16 - (-80 )= 96\).

The discriminant is a positive number but not a perfect square. Therefore, there are two irrational solutions.

\(\displaystyle \sqrt[3]{5-2x}+5=4\)

The first step to solving an equation where \(\displaystyle x\) is in a radical is to isolate the radical. To do this, we need to subtract the 5 from both sides.

\(\displaystyle \sqrt[3]{5-2x}+5-5=4-5\)

\(\displaystyle \sqrt[3]{5-2x}=-1\) 

Now that the radical is isolated, clear the radical by raising both sides to the power of 3. Note:

 \(\displaystyle (-1)^3 = -1 \cdot -1 \cdot -1 = -1\)

\(\displaystyle (\sqrt[3]{5-2x})^3=(-1)^3\)

\(\displaystyle {5-2x}=-1\)

Now we want to isolate the \(\displaystyle x\) term. First, subtract the 5 from both sides.

\(\displaystyle {5-2x}-5=-1-5\)

\(\displaystyle -2x=-6\)

Finally, divide both sides by \(\displaystyle -2\) to solve for \(\displaystyle x\).

\(\displaystyle \frac{-2x}{-2}=\frac{-6}{-2}\)

\(\displaystyle x=3\)

To solve for \(\displaystyle x\) in a literal equation, use the properties of algebra to isolate \(\displaystyle x\) on one side, just as if you were solving a regular equation. 

\(\displaystyle \frac{2x+3y }{c} = 20\)

Multiply both sides by \(\displaystyle c\):

\(\displaystyle \frac{2x+3y }{c} \cdot c = 20 \cdot c\)

\(\displaystyle 2x+3y = 20 c\)

Subtract \(\displaystyle 3y\) from both sides:

\(\displaystyle 2x+3y - 3y = 20 c - 3y\)

\(\displaystyle 2x = 20 c - 3y\)

Multiply both sides by \(\displaystyle \frac{1}{2}\), distributing on the right:

\(\displaystyle \frac{1}{2} \cdot 2x =\frac{1}{2} \cdot (20 c - 3y)\)

\(\displaystyle x =\frac{1}{2} \cdot 20 c -\frac{1}{2} \cdot 3y\)

\(\displaystyle x =10 c -\frac{3}{2} y\),

the correct response.

To solve for \(\displaystyle x\) in a literal equation, use the properties of algebra to isolate \(\displaystyle x\) on one side, just as if you were solving a regular equation. 

\(\displaystyle 9x^{2}+z = y\)

Subtract \(\displaystyle z\) from both sides:

\(\displaystyle 9x^{2}+z - z = y - z\)

\(\displaystyle 9x^{2} = y - z\)

Divide both sides by 9:

\(\displaystyle \frac{9x^{2}}{9} = \frac{y - z}{9}\)

\(\displaystyle x^{2} = \frac{y - z}{9}\)

Take the square root of both sides:

\(\displaystyle x =\sqrt{ \frac{y - z}{9}}\)

Simplify the expression on the right by splitting it, and taking the square root of numerator and denominator:

\(\displaystyle x = \frac{\sqrt {y - z}}{\sqrt {9}}\)

\(\displaystyle x = \frac{\sqrt {y - z}}{3}\),

the correct response.

To solve for \(\displaystyle x\) in a literal equation, use the properties of algebra to isolate \(\displaystyle x\) on one side, just as if you were solving a regular equation. 

First, take the reciprocal of both sides:

\(\displaystyle \frac{y+1}{x+1} = 3\)

\(\displaystyle \frac{x+1}{y+1} = \frac{1}{3}\)

Multiply both sides by \(\displaystyle y+1\):

\(\displaystyle \frac{x+1}{y+1} \cdot (y+1)= \frac{1}{3}(y+1)\)

\(\displaystyle x+1= \frac{1}{3}(y+1)\)

Distribute on the right:

\(\displaystyle x+1= \frac{1}{3} \cdot y+ \frac{1}{3} \cdot 1\)

\(\displaystyle x+1= \frac{1}{3} y+ \frac{1}{3}\)

Subtract 1 from both sides, rewriting 1 as \(\displaystyle \frac{3}{3}\) to facilitate subtraction:

\(\displaystyle x+1 - 1 = \frac{1}{3} y+ \frac{1}{3} - 1\)

\(\displaystyle x = \frac{1}{3} y+ \frac{1}{3} - \frac{3}{3}\)

\(\displaystyle x = \frac{1}{3} y - \frac{2}{3}\),

the correct response.